Question

Arrange the following ions in the order of decreasing X-O bond length where X is the centre
A. $Cl{{O}_{4}}^{-},S{{O}_{4}}^{2-},P{{O}_{4}}^{3-},Si{{O}_{4}}^{4-}$
B. $Si{{O}_{4}}^{4-},P{{O}_{4}}^{3-},S{{O}_{4}}^{2-},Cl{{O}_{4}}^{-}$
C. $Si{{O}_{4}}^{4-},P{{O}_{4}}^{3-},Cl{{O}_{4}}^{-},S{{O}_{4}}^{2-}$
D. $Si{{O}_{4}}^{4-},S{{O}_{4}}^{2-}P{{O}_{4}}^{3-},Cl{{O}_{4}}^{-}$

Answer 1

Hint: Bond length of any bond attached between the atoms in a molecule is the measure of the distance between the nuclei of the bonded atoms. Bond length depends on the bond order.

Complete step-by-step answer:
Bond length of the bonds in any molecule depends on varied factors. Some of them are:
- Size of the atom: A greater size atom has more length of bond.
- Multiplicity of a bond: Greater the multiplicity of the bonds less is the bond length. Like a compound with triple bond has less bond length, then the compound with double bonds, while a compound with single bonds has greatest bond length.
- Types of hybridization: S orbital has a small size, so, more is the s character, and less is the bond length.
- Bond order: single bonds or lesser bonds have larger bond length as compared to molecules with more bond order. Less bond order means more bond length and high bond order means less bond length.
Due to all these facts it can be inferred that, single bonds > double bonds > triple bonds. Single bonds have the highest bond length while triple bonds have the least bond length. Also the bond order values of $Si{{O}_{4}}^{4-},P{{O}_{4}}^{3-},S{{O}_{4}}^{2-},Cl{{O}_{4}}^{-}$ _{are 1, 1.25, 1.5, and 1.75 respectively.}
So, the order of the decreasing bond length becomes, $Si{{O}_{4}}^{4-}>P{{O}_{4}}^{3-}>S{{O}_{4}}^{2-}>Cl{{O}_{4}}^{-}$

Hence, the correct option is C.

Note: The order is due to the presence of no double bonds in $Si{{O}_{4}}^{4-}$, presence of 1 double bond in oxygen of $P{{O}_{4}}^{3-}$, presence of 2 double bond in oxygen of$S{{O}_{4}}^{2-}$, and the presence of 3 double bond in oxygen of$Cl{{O}_{4}}^{-}$.